Ecuatii si inecuatii de gradul al doilea si reductibile la gradul al ...

sau
Prin urmare
⎡
⎢
⎣
4 < x + 7
x
5 < x + 7
x
g) Se grupeaza t = x +
de unde rezulta
< 9
2 ,
< 8,
⎡
x > 0,
⎢ x < 0,
⎢ ⎧
⎢ ⎪⎨
x > 0,
⎢
⎢
⎣ x < 0,
⎪⎩
1 < x < 7,
de unde
⇔
⎡ ⎧
⎢ ⎪⎨ ⎢ ⎪⎩ ⎢ ⎧
⎢ ⎪⎨ ⎢
⎣
⎪⎩
x ∈ ∅,
x2 − 4x + 7
> 0,
x
2x2 − 9x + 14
< 0,
x
x2 − 5x + 7
> 0,
x
x2 − 8x + 7
< 0,
x
x ∈ (1; 7),
⇔ x ∈ (1; 7).
sau
5 + 3
, t = x + 4 (a se vedea (12)) si inecuatia devine
2
(t + 1) 4 + (t − 1) 4 < 272
t 4 + 6t 2 − 135 < 0
(t 2 − 9)(t 2 + 15) < 0 sau |t| < 3.
Prin urmare |x + 4| < 3. Se utilizeaza proprietatile modulului si se obtine
I. Sa se rezolve ecuatiile
1. x(x + 1)(x + 2)(x + 3) = 24.
2. (1 − x)(2 − x)(x + 3)(x + 3) = 84.
3. (x + 2) 4 + x 4 = 82.
|x + 4| < 3 ⇔ −3 < x + 4 < 3 ⇔ −7 < x < −1.
Exercitii pentru autoevaluare
4. (2x 2 + 5x − 4) 2 − 5x 2 (2x 2 + 5x − 4) + 6x 4 = 0.
x 2
x 2
5. + = 6.
x + 1 x − 1
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